Question

Without actually calculating the cubes, find the values of ${\left(28\right)}^3+{(-15)}^3+{(-13)}^3$

Answer 1

We know the corollary: if $a+b+c=0$ then $a^3+b^3+c^3=3abc$

Using the above corollary taking $a=28$, $b=-15$ and $c=-13$, we have $a+b+c=28-15-13=0$ then the value of $(28{)}^3+(-15{)}^3+(-13{)}^3$ is:

$(28{)}^3+(-15{)}^3+(-13{)}^3=3[28\times -15\times -13]=16380$

Hence, $(28{)}^3+(-15{)}^3+(-13{)}^3=16380$